I have a special use case where I want to discard all the contractions from the string and select only words followed by alphabets which do not contain any special character.
For eg:
string = "~ ASAP ASCII Achilles Ada Stackoverflow James I'd I'll I'm I've"
string.scan(/\b[A-z][a-z]+\b/)
#=> ["Achilles", "Ada", "Stackoverflow", "James", "ll", "ve"]
Note: It's not discarding the whole word I'll and I've
Can someone please help how to discard the whole word which contains contractions?
Try this Regex:
(?:(?<=\s)|(?<=^))[a-zA-Z]+(?=\s|$)
Explanation:
(?:(?<=\s)|(?<=^)) - finds the position immediately preceded by either start of the line or by a white-space
[a-zA-Z]+ - matches 1+ occurrences of a letter
(?=\s|$) - The substring matched above must be followed by either a whitespace or end of the line
Click for Demo
Update:
To make sure that not all the letters are in upper case, use the following regex:
(?:(?<=\s)|(?<=^))(?=\S*[a-z])[a-zA-Z]+(?=\s|$)
Click for Demo
The only thing added here is (?=\S*[a-z]) which means that there must be atleast one lowercase letter
I know that there's an accepted answer already, but I'd like to give my own shot:
(?<=\s|^)\w+[a-z]\w*
You can test it here. This regex is shorter and more efficient (157 steps against 315 from the accepted answer).
The explanation is rather simple:
(?<=\s|^)- This is a positive look behind. It means that we want strings preceded by a whitespace character or the start of the string.
\w+[a-z]\w* - This one means that we want strings composed by letters only (word characters) containing least one lowercase letter, thus discarding words which are whole uppercase. Along with the positive look behind, the whole regex ends up discarding words containing special characters.
NOTE: this regex won't take into account one-letter words. If you want to accomplish that, then you should use \w*[a-z]\w* instead, with a little efficiency cost.
Related
I'm trying to find a regex that will split a piece of text into sentences at ./?/! that is followed by a space that is followed by a capital letter.
"Hello there, my friend. In other words, i.e. what's up, man."
should split to:
Hello there, my friend| In other words, i.e. what's up, man|
I can get it to split on ./?/!, but I have no luck getting the space and capital letter criteria.
What I came up with:
.split("/. \s[A-Z]/")
Split a piece of text into sentences based on the criteria that it is a ./?/! that is followed by a space that is followed by a capital letter.
You may use a regex based on a lookahead:
s = "Hello there, my friend. In other words, i.e. what's up, man."
puts s.split(/[!?.](?=\s+\p{Lu})/)
See the Ruby demo. In case you also need to split with the punctuation at the end of the string, use /[!?.](?=(?:\s+\p{Lu})|\s*\z)/.
Details:
[!?.] - matches a !, ? or . that is...
(?=\s+\p{Lu}) - (a positive lookahead) followed with 1+ whitespaces followed with 1 uppercase letter immediately to the right of the current location.
See the Rubular demo.
NOTE: If you need to split regular English text into sentences, you should consider using existing NLP solutions/libraries. See:
Pragmatic Segmenter
srx-english
The latter is based on regex, and can easily be extended with more regular expressions.
Apart from Wiktor's Answer you can also use lookarounds to find zero width and split on it.
Regex: (?<=[.?!]\s)(?=[A-Z]) finds zero width preceded by either [.?!] and space and followed by an upper case letter.
s = "Hello there, my friend. In other words, i.e. what's up, man."
puts s.split(/(?<=[.?!]\s)(?=[A-Z])/)
Output
Hello there, my friend.
In other words, i.e. what's up, man.
Ruby Demo
Update: Based on Cary Swoveland's comment.
If the OP wanted to break the string into sentences I'd suggest (?<=[.?!])\s+(?=[A-Z]), as it removes spaces between sentences and permits the number of such spaces to be greater than one
I am trying to check if a string contains at least one lowercase letter, uppercase letter, and a number, but not punctuation (including spaces).
For example
4aBc8Fk3 should match
4aBc 8.;3 should not match
I tried the following, but it matches spaces:
^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9]).{6,}[^[:punct:]]$
Any ideas how to not match strings containing punctuation including spaces?
The regular expression you have got there does the following for as far as I understand (I'm not familiar with the ruby variety, and still quite new to regex myself; this will give you an idea, but may not be 100% correct):
Go to the beginning of the string
Ensure the string matches any number of any characters followed by a lowercase letter, e.g. --a
Ensure the string matches any number of any characters followed by an uppercase letter, e.g.--aA
Ensure the string matches any number of any characters followed by a number, e.g. --aA0
If that is all true, make sure the beginning of the string is followed by at least 6 random characters, e.g.--aA0-
Ensure that is followed by a single non-punctuation character (although this is the part I'm not sure about, as I haven't used character classes before, and don't know if it's [^[:punct:]] or [^:punct:]), e.g. --aA0-c
Ensure that is followed directly by the end of the string
Now, the lookaheads would also allow a different order of occurrences, e.g. 0---Aa, as long as the string contains any characters followed by what they are looking for.
What you probably want is ^[a-zA-Z0-9]{6,}$, i.e. at least six characters, with the characters being letters and numbers (though that would also allow aaaaaa, for example).
Maybe try ^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])[a-zA-Z0-9]{6,}$ to make sure each group is present, and to get alpha-numerical characters (at least six of them) only.
I always use a tool such as http://www.regexpal.com/ to slowly build up my regex and to see where I go wrong, deconstructing a "bad" regex until I get to a "good" one, then slowly adding to it again.
Hope that helps. :)
P.S.: I'm still a bit unclear how many characters you want to match in total, i.e. if the string is fixed length or not...?
I am new to ruby and I'm trying to work with regex.
I have a text which looks something like:
HEADING
Some text which is always non capitalized. Headings are always capitalized, followed by a space or nothing more.
YOU CAN HAVE MULTIPLE WORDS IN HEADING
I'm using this regular expression to choose all headings:
^[A-Z]{2,}\s?([A-Z]{2,}\s?)*$
However, it matches all headings which does not contain chars as Č, Š, Ž(slovenian characters).
So I'm guessing [A-Z] only matches ASCII characters? How could I get utf8?
You are right in that when you define the ASCII range A-Z, the match is made literally only for those characters. This is to do with the history of characters on computers, more and more characters have been added over time, and they are not always structured in an encoding in ways that are easy to use.
You could make a larger character class that matches the slovenian characters you need, by listing them.
But there is a shortcut. Someone else has already added necessary data to the Unicode data so that you can write shorter matches for "all uppercase characters": /[[:upper:]]/. See http://ruby-doc.org//core-2.1.4/Regexp.html for more.
Altering your regular expression with just this adjustment:
^[[:upper:]]{2,}\s?([[:upper:]]{2,}\s?)*$
You may need to adjust it further, for instance it would not match the heading "I AM A HEADING" due to the match insisting each word is at least two letters long.
Without seeing all your examples, I would probably simplify the group matching and just allow spaces anywhere:
^[[:upper:]\s]+$
You can use unicode upper case letter:
\p{Lu}
Your regex:
\b\p{Lu}{2,}(?:\s*\p{Lu}{2,})\b
RegEx Demo
I am using Notepad++ to check logs. I want to define custom syntax highlighting for timestamps and log levels. Highlighting logs levels works fine (defined as keywords). However, I am still struggling with highlighting timestamps of the form
06 Mar 2014 08:40:30,193
Any idea how to do that?
If you just want simple highlighting, you can use Notepad++'s regex search mode. Open the Find dialog, switch to the Mark tab, and make sure Regular Expression is set as the search mode. Assuming the timestamp is at the start of the line, this Regex should work for you:
^\d{2}\s[A-Za-z]+\s\d{4}\s\d{2}:\d{2}:\d{2},[\d]+
Breaking it down bit by bit:
^ means the following Regex should be anchored to the start of the line. If your timestamp appears anywhere but the start of a line, delete this.
\d means match any digit (0-9). {n} is a qualifier that means to match the preceding bit of Regex exactly n times, so \d{2} means match exactly two digits.
\s means match any whitespace character.
[A-Za-z] means match any character in the set A-Z or the set a-z, and the + is a qualifier that means match the preceding bit of Regex 1 or more times. So we're looking for an alphabetic character sequence containing one or more alphabetic characters.
\s means match any whitespace character.
\d{4} is just like \d{2} earlier, only now we're matching exactly 4 digits.
\s means match any whitespace character.
\d{2} means match exactly two digits.
: matches a colon.
\d{2} matches exactly two digits.
: matches another colon.
\d{2} matches another two digits.
, matches a comma.
[\d]+ works similarly to the alphabetic search sequence we set up earlier, only this one's for digits. This finds one or more digits.
When you run this Regex on your document, the Mark feature will highlight anything that matches it. Unlike the temporary highlighting the "Find All in Document" search type can give you, Mark highlighting lasts even after you click somewhere else in the document.
I need to filter all lines with words starting with a letter followed by zero or more letters or numbers, but no special characters (basically names which could be used for c++ variable).
egrep '^[a-zA-Z][a-zA-Z0-9]*'
This works fine for words such as "a", "ab10", but it also includes words like "b.b". I understand that * at the end of expression is problem. If I replace * with + (one or more) it skips the words which contain one letter only, so it doesn't help.
EDIT:
I should be more precise. I want to find lines with any number of possible words as described above. Here is an example:
int = 5;
cout << "hello";
//some comments
In that case it should print all of the lines above as they all include at least one word which fits the described conditions, and line does not have to began with letter.
Your solution will look roughly like this example. In this case, the regex requires that the "word" be preceded by space or start-of-line and then followed by space or end-of-line. You will need to modify the boundary requirements (the parenthesized stuff) as needed.
'(^| )[a-zA-Z][a-zA-Z0-9]*( |$)'
Assuming the line ends after the word:
'^[a-zA-Z][a-zA-Z0-9]+|^[a-zA-Z]$'
You have to add something to it. It might be that the rest of it can be white spaces or you can just append the end of line.(AFAIR it was $ )
Your problem lies in the ^ and $ anchors that match the start and end of the line respectively. You want the line to match if it does contain a word, getting rid of the anchors does what you want:
egrep '[a-zA-Z][a-zA-Z0-9]+'
Note the + matches words of length 2 and higher, a * in that place would signel chars too.